Mole & Stoichiometry Practice Homework KEY
Due: Monday 1/7/13 (Red) or Tuesday 1/8/13 (Blue)
1. How many particles (molecules) are in 4.65 moles of Nitrogen Dioxide (NO2)?
4.65 mol NO2
x
46.01 g NO2 = 214 g NO2
1 mol NO2
2. What is the molar mass of Sodium Sulfate (Na2SO4)?
Na (2 x 22.99 g) + S (1 x 32.07 g)
+ O (4 x 16.00 g) = 142.05 g Na2SO4
3. How many moles of Chlorine gas are in a container that is 3.25 liters, if at STP?
3.25 L Cl2
x
1 mol Cl2
=
0.145 mol Cl2
22.4 L Cl2
4. What is the percent (%) composition by mass for each element in the butanol (C3H7OH)
molecule?
C
3 x 12.01
=
36.03 g C
36.03 g C/ 60.11 g total x 100 = 59.94 % C
H
8 x 1.01
=
8.08 g H
8.08 g H/ 60.11 g total x 100 = 13.4 % H
=
16.00 g O
60.11 g C3H7OH
16.00 g O/ 60.11 g total x 100 = 26.62 % O
O
1
x 16.00
5. What is the mass of CO2 gas that is produced by a decomposition reaction
(CaCO3 
CaO +
0.0345 L CO2
x
CO2) if a student collects 0.0345 Liters of gas at STP?
1 mol CO2
22.4 L CO2
x
44.01 g CO2 =
1 mol CO2
0.0678 g CO2
6. Chemical analysis of Succinic acid indicates it is composed of 40.68% carbon, 5.08%
hydrogen, and 52.24% oxygen. Determine the empirical formula.
% Given
40.68 % C
Assume 100 g
40.68 g C
5.08 % H
5.08 g H
52.24 % O
52.24 g O
Convert to Moles
x 1 mol C
12.01 g C
x 1 mol H
1.01 g H
x 1 mol O
16.00 g O
Mole Answer
= 3.387 mol C
= 5.03 mol H
= 3.265 mol O
Ratio***
3.387 mol = 1
3.265 mol
5.03mol = 1.5
3.265 mol
3.265 mol = 1
3.265 mol
*** The ratio is achieved by dividing the smallest number of moles into itself and all other mole values.
This ratio is not a whole number C1H1.5O1
So you must double all subscripts to achieve a whole number ratio of atoms, this leads to C2H3O2
7. If the molar mass of Succinic acid is 118.1 g/mol, what is its molecular formula?
Molar Mass
= 118.1 g/mol
Empirical Mass
=
2
so double the subscripts
C4H6O4
59.05 g/mol
8. When 104.5 g of Iron (III) Oxide reacts with an excess or carbon monoxide during a
laboratory experiment and 54.6 g of Iron is collected by the lab student according to the
following equation:
Fe2O3 (s)
+
3 CO
(g)
→
2 Fe
(s)
+
3 CO2 (g)
What is the percent yield of the experiment?
104.5 g Fe2O3
x
1 mol Fe2O3
159.70 g Fe2O3
Experimental x 100 = % Yield
Theoretical
x 2 mol Fe
x
55.85 g Fe = 73.09 g Fe
1 mol Fe2O3
1 mol Fe
54.6 g Fe x 100 = 74.7 % Yield
73.09 g Fe
9. Consider the following reaction: 3 NH4NO3 + Na3PO4  (NH4)3PO4 + 3 NaNO3
We have 30.00 grams of ammonium nitrate and 50.00 grams of sodium phosphate determine the
following (use dimensional analysis to prove your answer):
What is the maximum amount of Ammonium Phosphate that can be formed?
30.00 g (NH4)(NO3) x 1 mol (NH4)(NO3) x 1 mol (NH4)3(PO4) x 149.12 g (NH4)3(PO4) =
80.06 g (NH4)(NO3) 3 mol (NH4)(NO3)
1 mol (NH4)3(PO4)
= 18.63 g (NH4)3(PO4)
50.00 g Na3PO4 x 1 mol Na3PO4 x 1 mol (NH4)3(PO4) x 149.12 g (NH4)3(PO4) =
163.94 g Na3PO4 1 mol Na3PO4
1 mol (NH4)3(PO4)
= 45.48 g (NH4)3(PO4)
What is the limiting reagent? (NH4)(NO3) What is the excess reagent? Na3(PO4)